After the end of the first round of my program (Collatz function), I want my program to continue to calculate from the previous result.
I wrote a program with the possibility of one iteration, which starts from the input:
def collatz(number):
if number % 2 == 0: #parity conditions value
return number // 2
if number % 2 == 1: #parity oddness value
return 3 * number + 1
result = 5
while True:
print ('Type your number')
result = int(input())
print (collatz(result))
If you want to re-run the function over and over again with the last ran result - store the return value in variable and call the function with it.
def collatz(number):
if number % 2 == 0: #parity conditions value
return number // 2
if number % 2 == 1: #parity oddness value
return 3 * number + 1
print ('Type your number')
result = int(input()) #first time the input will come from the user
while True:
result = collatz(result) #calculate new result
print (result)
#if you want to add break out of the loop put it here
Related
I'm new to python and I'm trying to create something that if the number is odd for example 7 it does this (7 * 3+1) and if it is even for example 6 it does this (6/2) but the problem is that I want to loop this but it updates the number every output from the example earlier 7 I want this to turn to (22/2) and so on and if the number 1 is reached it stops.
output = []
number = 7
def mat(self):
if (self % 2) == 0:
even = self / 2
return even
else:
odd = self * 3 + 1
return odd
while mat(number) != 1:
output.append(mat(number))
output.append(mat(mat(number))
print(output)
this part doesn't work because it will never reach 1 and it only has 1 output (22) starting from the number 7 :
while mat(number) != 1:
output.append(mat(number))
output.append(mat(mat(number))
To update number, you need to assign it:
number = mat(number)
The best place to do this is in the while loop:
while number != 1:
number = mat(number)
For an exercise like this, it makes sense to just print the value on each iteration rather than trying to create an array of results:
while number != 1:
print(number)
number = mat(number)
Just update the value
For while loop:
a = 0
while a<10:
print("Hello World")
a = a + 1
For for loop:
a = 0
for i in range(10):
print("Hello World")
a = a + 1
if a >= 10:
break
I've got an assignment which requires me to use a Python recursive function to output the factors of a user inputted number in the form of below:
Enter an integer: 6 <-- user input
The factors of 6 are:
1
2
3
6
I feel like a bit lost now and have tried doing everything myself for the past 2 hours but simply cannot get there. I'd rather be pushed in the right direction if possible than shown where my code needs to be changed as I'd like to learn
Below is my code:
def NumFactors(x):
for i in range(1, x + 1):
if x == 1:
return 1
if x % i == 0:
return i
return NumFactors(x-1)
x = int(input('Enter an integer: '))
print('The factors of', x, 'are: ', NumFactors(x))
In your code the problem is the for loop inside the method. The loop starts from one and goes to the first if condition and everything terminates there. That is why it only prints 1 as the output this is a slightly modified version of your own code. This should help. If you have any queries feel free to ask.
def factors(x):
if x == 1:
print(1 ,end =" ")
elif num % x == 0:
factors(x-1)
print(x, end =" ")
else:
factors(x-1)
x = num = int(input('Enter an integer: '))
print('The factors of', x, 'are: ',end =" ")
factors(x)
Since this question is almost 3 years old, I'll just give the answer rather than the requested push in the right direction:
def factors (x,c=1):
if c == x: return x
else:
if x%c == 0: print(c)
return factors(x,c+1)
Your recursion is passing down x-1 which will not give you the right value. For example: the number of factors in 6 cannot be obtained from the number of factors in 5.
I'm assuming that you are not looking for the number of prime factors but only the factors that correspond to the multiplication of two numbers.
This would not normally require recursion so you can decide on any F(n) = F(n-1) pattern. For example, you could use the current factor as a starting point for finding the next one:
def NumFactors(N,F=1):
count = 1 if N%F == 0 else 0
if F == N : return count
return count + NumFactors(N,F+1)
You could also optimize this to count two factors at a time up to the square root of N and greatly reduce the number of recursions:
def NumFactors(N,F=1):
count = 1 if N%F == 0 else 0
if N != F : count = count * 2
if F*F >= N : return count
return count + NumFactors(N,F+1)
Please help, I cannot figure out why this code does not work. I think the first loop runs forever but I don't know why!
def NTN():
list1 = []
count = 0
number = 0
Start = input('Type Start Number')
while number != Start:
count = count + 1
number = number + count
Stop = input('Type Stop Number')
while number != Stop:
count = count + 1
number = number + count
if number != Stop:
(list1).append(number)
return (list1)
print(NTN())
You are increasing number by increasing amounts in every iteration. Here's an idea of how it is increasing. Assume Start = 4
After 1 loop, count = 1 and number = 1, increase of 1
After 2 loops, count = 2 and number = 3, increase of 2
After 3 loops, count = 3 and number = 6, increase of 3
Since number is never really equal to 4, the loop never ends. What you need probably is while number <= Start. That would terminate the loop after 3 iterations when number is past 4.
change "number != Start" and "number != Stop" to "number < Start" and "number < Stop" in all places, and it should work.
What went wrong: if Start is 2, then in the first iteration of the while loop, count becomes 0+1=1 and number becomes 0+1=1; in the second iteration, count becomes 1+1=2 and number becomes 1+2=3, which bypasses 2. Since your while loop only ends when number is equal to Start, it never ends.
A couple of side-points:
By convention Python variable and function names are lower-case.
input() returns a string; if you want a number you have to convert it ie with int() or float(). (Note: if you are using Python 2.x input() calls eval() which is really awful design - you should be using int(raw_input()) instead.)
so,
# This code assumes Python 3.x
from math import ceil, sqrt
def get_int(prompt):
"""
Prompt until an integer value is entered
"""
while True:
try:
return int(input(prompt))
except ValueError:
print("Please enter an integer!")
def tri(n):
"""
Return triangular number n,
ie the sum of (1 + 2 + ... + n)
"""
# using Gaussian sum
return n * (n + 1) // 2
def reverse_tri(t):
"""
For positive integer t,
return the least positive integer n
such that t <= tri(n)
"""
# derived by quadratic formula from
# n * (n + 1) // 2 >= t
return int(ceil(((sqrt(8 * t + 1) - 1) / 2)))
def ntn(start, stop):
"""
Return a list of triangular numbers
such that start <= tri < stop
"""
a = reverse_tri(start)
b = reverse_tri(stop)
return [tri(n) for n in range(a, b)]
def main():
start = get_int('Enter a start number: ')
stop = get_int('Enter a stop number: ')
lst = ntn(start, stop + 1) # include stop number in output
print(lst)
if __name__ == "__main__":
main()
An Emirp is a prime number whose reversal is also a prime. For
example, 17 is a prime and 71 is a prime, so 17 and 71 are emirps.
Write a program that prints out the first N emirps, five on each line.
Calculate the first N emirp (prime, spelled backwards) numbers, where
N is a positive number that the user provides as input.
Implementation Details
You are required to make use of 2 functions (which you must write).
isPrime(value) # Returns true if value is a prime number. reverse
(value) # Returns the reverse of the value (i.e. if value is 35,
returns 53). You should use these functions in conjunction with logic
in your main part of the program, to perform the computation of N
emirps and print them out according to the screenshot below.
The general outline for your program would be as follows:
Step 1: Ask user for positive number (input validation) Step 2:
Initialize a variable Test to 2 Step 3: While # emirps found is less
than the input:
Call isPrime with Test, and call it again with reverse(Test).
If both are prime, print and increment number of emirps found. Test++ Hint - to reverse the number, turn it into a string and then
reverse the string. Then turn it back into an int!
MY CODE:
n = 0
count = 0
i = 1
def isPrime(value):
test = 2
count = 0
while(test < value):
if( value % test == 0):
count+=count
test+=test
if(count == 0):
return 1
else:
return 0
def reverse(value):
reverse = 0
while(value > 0):
reverse = reverse * 10 + (value % 10)
value = value / 10
return reverse
n = float(input("Please enter a positive number: "))
while(count < n):
i+=i;
if(isPrime(i)):
if(isPrime(reverse(i))):
print("i" + "\n")
count+=count
if((count % (5)) == 0 ):
print("\n")
where are the mistakes?
UPDATED CODE BUT STILL NOT RUNNING:
n = 0
count = 0
i = 1
def isPrime(value):
test = 2
while(test < value):
if( value % test != 0):
test +=1
else:
return 0
def reverse(value):
return int(str(value)[::-1])
i = int(input("Please enter a positive number: "))
count = 0
while(count < 5):
if(isPrime(i)):
if(isPrime(reverse(i))):
print(str(i) + "\n")
count += 1
i += 1
else:
i += 1
else:
i +=1
There are a lot of issues with your code. I altered the function isPrime. There is among other no need for count here and if you want to increment test by 1 every loop use test +=1:
def isPrime(value):
test = 2
while(test < value):
if( value % test != 0):
test +=1
else:
return 0
return 1
There is an easy way to reverse digits by making value a string in reversed order and to convert this into an integer:
def reverse(value):
return int(str(value)[::-1])
You among other have to assign the input to i. n in your code is a constant 5. You have to increment i by one in any condition and increment the count by one if i is an emirp only:
i = int(input("Please enter a positive number: "))
count = 0
while(count < 5):
if(isPrime(i)):
if(isPrime(reverse(i))):
print(str(i) + "\n")
count += 1
i += 1
else:
i += 1
else:
i +=1
def emrip_no(num):
i=0
j=0
for i in range(1,num+1):
a=0
for j in range(1,i+1):
if(i%j==0):
a+=1
if(a==2):
print("Number is a prime number")
k=0
l=0
rv=0
while(num!=0):
r=num%10
rv=(rv*10)+r
num=num//10
print("It's reverse is: ",rv)
for k in range(1,rv+1):
b=0
for l in range(1,k+1):
if(k%l==0):
b+=1
if(b==2):
print("It's reverse is also a prime number")
else:
print("It's reverse is not a prime number")
n=int(input("Enter a number: "))
emrip_no(n)
I have a while loop which calls the function mainrt() on each iteration.
if __name__ == "__main__":
inp = sys.stdin.read()
inpList = inp.split('\n')
inpList.pop()
for n in inpList:
i = 0
p = 0
n = int (n)
while True:
i += 1
p = n*i
if n == 0:
print "INSOMNIA"
break
else:
res = True
res = mainrt(p)
if res == False:
print p
break
And the mainrt()
def mainrt(n):
#print n
while True:
rem = n % 10
if rem in diMon:
pass
else:
diMon.insert(rem,rem)
if len(diMon) == 10:
return False
break
n = n/10
if n == 0:
return True
break
else:
continue
The problem is as i take input from stdin.read() the first line of the input processed properly by the function, but the second line of the input get printed as it is. It is not being processed by the function
example
INPUT
3
5
OUTPUT SHOLD BE
30
90
But instead I get
30
5
Why the function not processing the input the second time???
There is no run time error so far.
In your mainrt function I do not see that you declare diMon list, so it looks like it is a global variable and you do not clean that list. That mean your mainrt return False at the first check of if len(diMon) == 10: for the second input. You should declare diMon at the beginning od mainrt function or clear it at the end of while loop body.
EDIT:
Now I checked your code one more time and I suggest you to declare diMon at the beginning of for loop
for n in inpList:
diMon = []
i = 0
p = 0
n = int (n)